EXPLANATION OF NEON IONIZATIONS
By Prof. L. Kaliambos (Natural Philosopher in New Energy) May 10, 2015 Neon is a chemical element with symbol Ne and atomic number 10. However despite the enormous success of the Bohr model and the quantum mechanics of Schrodinger in explaining the principal features of the hydrogen spectrum and of other one-electron atomic systems, so far neither was able to provide a satisfactory explanation of ionizations of many electon atoms related to the chemical properties of atoms. Though such properties were modified by the periodic table initially proposed by the Russian chemist Mendeleev the reason of this subject of ionizations of elements remained obscure under the influence of the invalid theory of special relativity. (EXPERIMENTS REJECT RELATIVITY). It is of interest to note that the discovery of the electron spin by Uhlenbeck and Goudsmit (1925) showed that the peripheral velocity of a spinning electron is greater than the speed of light, which is responsible for understanding the electromagnetic interaction of two electrons of opposite spin. So it was my paper “Spin-spin interactions of electrons and also of nucleons create atomic molecular and nuclear structures” (2008), which supplied the clue that resolved this puzzle. Under this condition we may use this image of Neon including the following ground state electron configuration: 1s2.2s2.2px2.2py2.2pz2. According to the “Ionization energies of the elements-WIKIPEDIA” the ionization energies (in eV) are the following: E1 = 21.5646 , E2 = 40.96328 , E3 = 63.45 , E4 = 97.12 , E5 = 126.21 , E6 = 157.93, E7 = 207.2759 , E8 = 239.0989 . E9 = 1195.8286 , and E10 = 1362.1995. See also my papers about the explanation of ionization energies of elements in my FUNDAMENTAL PHYSICS CONCEPTS and my paper of 2008. For simplicity we start with the E6 , E5, and E4 , while the E3 , E2 , and E1 are explained in the second chapter. EXPLANATION OF E6 = 157.93 eV E5 = 126.21 eV, AND E4 = 97.12 eV ''' Note that the total sumation of the above energies represents the total binding energy of 2px1 , 2py1 , and 2pz1 of charge (- 3e) . So under a perfect screening of the spherical shells 1s2 and 2s2, we should determine the effective ζ = 6, because +10e -2e -2e = +6e. However the deformation of spherical shells leads to ζ > 6 . For simplicity we start with the E6 = 157.93 eV = - E(2pz1). Of course in the absence of 5 electrons outside the 2s2 one gets the E(2pz1) which is the binding energy of the one electron with n = 2 based on the Bohr model. In this case of n = 2 we must find the effective ζ1 > 6, because the 1s2 and 2s2 of the charge (-4e) screen the charge (+10e ) of the nucleus. Of course for a perfect screening due to the spherical shells of 1s2 and 2s2 we should use ζ = 6 as E (2pz1) = (-13.6) ζ2/22 = (-13.6)62/22 = - 122.4 eV However the 2pz1 penetrates the 2s2 which leads to the great deformations not only of spherical shells 1s2 and 2s2 but also of 2pz1. Thus, in fact, writing E6 = 157.93 eV = - E(2pz1) = - (- 13.6) ζ12/22 we get ζ1 = 6.82 > 6 . Then adding the second electron 2py1 one gets the binding energy E (2pz1 + 2py1) = - (E6 + E5 ). Thus applying the Bohr model we write E(2pz1 + 2py1) = 2(-13.6057)ζ22/22 = - (E6 + E5 ) = - 284.14 eV Therefore one gets ζ2 = 6.464 > 6 Then adding the third electron ( 2px1) one gets the binding energy of 3 electrons as E (2pz1 + 2py1 + 2px1) = 3(-13.6) ζ32 /22 = - (E6 + E5 + E4 ) = - 381.26 Therefore we get ζ3 = 6 Here ζ3 = 6 means that the three electrons with parallel spin (S = 1) exert both electric and magnetic repulsions and exist at symmetrical positions providing a perfect screening . '''EXPLANATION OF E3 = 63.45 eV, E2 = 40.96328 eV AND E1 = 21.5646 eV BASED ON MY FORMULA OF 2008 ''' Adding the three more electrons of the ionization energies E3 , E2 , and E1 we get the total binding energy E (2px2 + 2py2 + 2pz2) = 3ζ2 +(16.95) ζ - 4.1)/22 = - ( E1 + E2 + E3 + E4 + E5 + E6 ) Since ( E1 + E2 + E3 + E4 + E5 + E6) = 507.24 eV the above equation can be written as 20.4 ζ2 - 12.7125ζ - 504.165 = 0 Then solving for ζ one gets ζ = 5.29 < 6. As in the cases of oxygen and fluorine ( see my papers in my "Fundamenta Physics Concepts") since ζ < 6 cannot exist we suggest that the correct ζ = 6 of a perfect screening gives n >2 . Under this condition we find that n = 2.28 . It means that the three orbitals of paired electrons do not lead to the deformations of 1s2 and 2s2 but differ from the symmetry of (2px1+ 2py1 +2pz1), because they do not exert magnetic repulsions. Here electric repulsions between the paired electrons of 2px2, 2py2 and 2pz2 lead to a perfect screening under n = 2.28 . Note that the two electrons of opposite spin (say the 2px2) do not provide any mutual repulsion because I discovered in 2008 that at very short inter-electron separations the magnetic attraction is stronger than the electric repulsion and leads to a vibration energy. However in the absence of a detailed knowledge about the mutual electromagnetic interaction between the electrons of the 2px2 or 2py2 or 2pz2 pairs today many physicists believe that it is due to the Coulomb repulsion between the two electrons of opposite spin. Under such fallacious ideas I published my paper of 2008. '''EXPLANATION OF E8 = 239.0989 eV, AND E7 = 207.2759 eV For simplicity we start with the E8 = -E( 2s1). According to the quantum mechanics the one electron (2s1) penetrates the 1s2 shell. Thus it leads to the deformations of both 1s2 and 2s1 spherical shells giving an effective ζ > 8, because the charge (- 2e) of the two electrons of 1s2 screens the charge (+10e) of nucleus. Since n = 2 we may write E8 = 239.0989 eV = -E(2s1) = - (-13.6057)ζ2/22 Therefore one gets ζ = 8.38 > 8 Then adding the second electron of opposite spin we get the binding energy, E(2s2), of the two electrons of opposite spin for n = 2 given by my formula of 2008: E(2s2) = + (16.95)ζ - 4.1 / 22 = - (E7 + E8 ) Since ( E7 + E8) = 446.3748 eV, the above equation could be written as 6.8025ζ2 - 4.2375ζ - 445.3498 = 0 Then, solving for ζ we get ζ = 8.4 > 8 Here 8.4 > 8.38 means that the repulsion of two electrons (2s2-1s2 ) is a little greater than the repulsion of one electron ( 2s1-1 s2 ) because the two electrons (2s2) of opposite spin behave like one particle. Note that In both cases the repulsions are due to only electric forces of the Coulomb law. Whereas in the case of the three electrons of 2px1, 2py1, and 2pz1 of parallel spin (S = 1) we observe a perfect screening, because the three electrons interact with both electric and magnetic repulsions from symmetrical positions. EXPLANATION OF Ε10 = 1362.1995 eV AND E9 = 1195.8286 eV As in the case of the helium the ionization energy E10 = 1362.1995 eV = - E(1s1) is due to the one remaining electron of 1s1 with n = 1. Thus we expect to calculate it by applying the simple Bohr model for Z = 10 as E10 = - (-13.6057 )Z2 /12 = - (-13.6057)10 2 /12 = 1360.57 eV. Surprisingly one sees here that after the ionizations the Bohr model gives the value of 1360.57 eV which is smaller than the experimental value of E10 = 1362.1995. Under this condition of ionizations I suggest that n = 1 becomes n < 1 due to the fact that the ionizations reduce the electron charges and now the nuclear charge is much greater than the electron charge of the remaining electron. So we may write E10 = 1362.1995 eV = (13.6057) Z2/n2 = (13.6057)102/n2 Then solving for n we get n = 0.9994. In the same way for calculating the E9 = 1195.8286 eV we must determine the simple quantum number n < 1 by applying my formula of 2008 as E9 = 1195.8286 eV = - E10 - E(1s2) = - 1362.1995 - [ (-27.21)102 + (16.95)10 - 4.1 ] / n2 Then solving for n we get n = 0.9995. Here 0.9995 > 0.9994, because the second electron increase the electron charge with respect to nuclear charge. Category:Fundamental physics concepts